Problem: $\text E = \left[\begin{array}{rr}5 & 3 \\ -2 & 1 \\ 4 & 1\end{array}\right]$ and $\text D = \left[\begin{array}{rr}-2 & -1 \\ 5 & 0\end{array}\right]$ Let $\text {H = ED}$. Find $\text H$. $ {H = }$
The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{E}$ and the first column of $\text{D}$. $ \text {H}=\left[\begin{array}{rr}{5} & {3} \\ -2 & 1 \\ 4 & 1\end{array}\right]\left[\begin{array}{rr} {-2} & -1 \\ {5} & 0\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(5,3)\cdot(-2,5)\\\\ &=5 \cdot -2 + 3\cdot 5\\\\ &=5 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $5 \cdot -1 + 3\cdot 0 = -5$ (Choice B) B $-2 \cdot 2 - 1\cdot 0 = -4$ (Choice C) C $-2 \cdot -2 + 1\cdot 5 = 9$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}5 & -5 \\ 9 & 2 \\ -3 & -4\end{array}\right]$